Validation Problem on Torsion


	Applies To

	Product(s):	STAAD.Pro
	Version(s):	All
	Environment:	N/A
	Area:	EN 1993-1-1:2005 (EC3)
	Subarea:	Validation Problem on Torsion
	Original Author:	Shayan Roy Bentley Technical Support Group

1. Introduction:

The following is a validation problem of a detailed torsion check including the warping stresses on a UB 203x203x60 which has been subjected to a uniformly distributed load across the full span and a torsional moment at the centre of the span.

The detailed torsion check in the EN 1993-1-1:2005 implementation in STAAD.Pro has been invoked by using the TOR 2 parameter.

The combination of the uniformly distributed load and the torsional moment resembles the load acting at a distance from the shear center.

The implementation calculates the torsional stresses and capacities as per P057. Please refer to STAAD.Pro EN 1993-1-1:2005 implementation: Calculation of Torsional Capacities of Cross Sections as per SCI P057 for details on this.

2 . Validation Problem:

A British I section beam (UB 203x203x60, S275 grade steel) , 4 m long fixed at both the ends has been subjected to a uniformly distributed load of 1 KN/m , a concentrated point load 1000 KN at the midpoint and a concentrated torque of 7.5 KNm at the midpoint.

The maximum Bending Moment (Mz) at the end is 51.33 KNm and at mid point is 50.7 KNm.

Maximum Shear Force (VY) is 52KN is at the supports.

3.1 . Text Input Information.

The following is the text input information of the Staad model:

STAAD SPACE

START JOB INFORMATION

ENGINEER DATE 07-Oct-13

END JOB INFORMATION

INPUT WIDTH 79

UNIT METER KN

JOINT COORDINATES

1 0 0 0; 2 4 0 0;

MEMBER INCIDENCES

1 1 2;

DEFINE MATERIAL START

ISOTROPIC STEEL

E 2.05e+008

POISSON 0.3

DENSITY 76.8195

ALPHA 1.2e-005

DAMP 0.03

TYPE STEEL

STRENGTH FY 253200 FU 407800 RY 1.5 RT 1.2

END DEFINE MATERIAL

MEMBER PROPERTY BRITISH

1 TABLE ST UC203X203X60

CONSTANTS

MATERIAL STEEL ALL

SUPPORTS

1 2 FIXED

LOAD 1 LOADTYPE None TITLE LOAD CASE 1

MEMBER LOAD

1 CMOM GX 7.5 2

1 CON GY -100 2

1 UNI GY -1

PERFORM ANALYSIS

PRINT MEMBER PROPERTIES

PARAMETER 1

CODE EN 1993-1-1:2005

SGR 1 ALL

CMT 3 ALL

TORSION 2 ALL

TRACK 2 ALL

CHECK CODE ALL

FINISH

3.2 . Manual Calculation:

Section: UB 203x203x60

Sectional Properties as per Bluebook

A = 76.4 cm²; D = 209.6 mm ; b = 205.8 mm ; t _f = 14.2 mm; t _w = 9.2 mm; r = 10.2 mm

Zz = 584 cm² ; Zy = 201 cm² ; Iz = 6120 cm⁴ ; Iy = 2060 cm⁴

J = 47.2 cm⁴; H = 1.97x10¹¹ mm⁶ ; G = 78846.15

Hence, torsional bending constant, a = sqrt (EH/GJ) =sqrt (205000*1.97*10¹¹ / 78846.15*47.2*10⁴ ) = 1041.72 mm.

A. Calculation of Torsional Properties of the section:

Normalized Warping Function

By Appendix A, P075, pg 112 ,

W_no = hB/4 = (209.6-14.2)*205.8 / 4 = 10053.3 mm².

Warping Statical Moment

S_w1 = hB²T/16 = (195.4*205.8^2*14.2)/6 = 7344862.36 mm⁴

(In Staad, a member is designed at 13 intermediate sections along the length of the member and the most critical section is reported. For the calculation purpose, we will only design only at z =1m and z =2m.)

B. Validation of the ratio reported for Clause 6.2.7(1):

The critical ratio for this clause is evaluated at 1m from start of the member as per the STAAD Output File.

As the ends are fixed for torsion and warping and the member is subjected to a concentrated torque, we will use Case 5 of Appendix B, P057 (as directed in Table 6.1, P057). This condition is specified by specifying the CMT parameter as 3.

z =1000, a = 1041.72, α= 0.5, αL/a = 1.9199, z/a = 0.9599, L/a = 3.8398

K₁= {(1/0.999*(1-3.483)+1/23.297*(3.483-1)+3.3368-1.9199)} / {1/23.247*(23.269+3.483*23.269-3.483-1)-1.9199-3.3368}

= 0.9989

K₂= 1/0.9989*1/23.247*(3.483-1)+(3.483-23.269+3.8398*23.247)/23.247

= 3.0955

K₃= 0.043+3.3368- 3.483/0.999

= -0.1067

K₄= 3.3368-3.4867+1.009

= 0.851

K₅ = 1/0.99891*1/0.999*(1-3.483)+(1-3.483*23.269)/23.247

= -5.931

Now,

Ф’ = (7.5*10⁶) / (78846.15*47.2*10⁴ )*1/1.9989*(0.744*1.1143-1.497+1)

= 3.3476*10^-5

Ф’’’ = (7.5*10⁶)/(78846.15*47.2*1041.72² *10⁴ )*1/1.9989*(0.744*1.1143-1.497)

= - 6*10^-11

As per equation 2.2 of P057,

Pure torsion component: T_p = GJ Ф’ = 78846.15 * 47.2*10⁴ * 3.3476*10^-5 / 10⁶= 1.25KNm (as reported by Staad)

Warping Torsion component: T_w = -EH Ф’’’ = 205000*1.97*10¹¹* -6*10^-11 / 10⁶= -2.42KNm (against the value reported as 2.5 by STAAD)

Pure Torsion Capacity:

T_{t, Rd}= fy/√3 *J/t = 275/√3*(47.4*10⁴)/14.2*1/10⁶

= 5.3 KNm (as reported by Staad)

Warping Torsion Capacity:

T_{w, Rd}= ( t*b^2*fy)/6 = ( 14.2*205.8²*275)/6 1/10⁶

= 27.6KNm (as reported by Staad).

Torsional Resistance T_Rd = T_{t, Rd}+ T_{w, Rd}

Now, T_p/(T_t,Rd) = 1.25/5 = 0.236

And, (T_w )/(T_w,Rd) = 2.42/27.6 = 0.08.

Hence, the critical ratio of 0.236 is reported by Staad for this clause. An extract from code check details of the STAAD Output file is as below:

C. Validation of the ratio calculated for clause 6.2.7(5):

Clause 6.2.7(5) refers back to the yield criterion of clause 6.2.1(5). The critical ratio for this clause is reported at a distance of 2 metres from the start of the member.

Torsion at section z=2m

z =2000, a = 1041.72, α= 0.5, αL/a = 1.9199, z/a = 1.9199, L/a = 3.8398

ф = (7.5*10⁶*1041.72)/(78846.15*47.2*10⁴ )*1/(1+0.9989)*[(0.9989*(-.1067)+0.851)*(2.483)-3.3368+1.9199

= 0.045

Ф’ = (7.5*10⁶)/(78846.15*47.2*10⁴ )*1/1.9989*(0.744*3.3368-3.483+1)

= -4.243*10^-8

Ф’’ = (7.5*10⁶)/(78846.15*47.2*1041.72*10⁴ )*1/1.9989*(0.744*3.483-3.3368)

= - 7.2146*10^-8

Ф’’’ = (7.5*10⁶)/(78846.15*47.2*1041.72² *10⁴ )*1/1.9989*(0.744*3.3368-3.483)

= -9.295*10^-11

As per equation 2.2 of P057,

Pure torsion component: T_p = GJ Ф’ = 78846.15 * 47.2*10⁴ * 0.045 / 10⁶= 0.0016

Warping Torsion component: T_w = -EH Ф’’’ = 205000*1.97*10¹¹* -9.295*10^-11 / 10⁶= -3.75KNm

Stress Check at 2m, Check for Elastic Verification as per CL 6.2.1.(5)

The program calculates the stress at 4 points of any cross sections as illustrated:

Let us calculate the stress at mid span, z = 2m from the start of the member.

Point 1:

Bending Stress:

σ_bz= M_z/Z_z = (50.7*10⁶)/(584*10³ ) (Z_z= I_z/(y/2)

= 86.82 N/mm²

σ_byt= M_yt/Z_y = Ф* M_z/Z_y ( CL 2.3, P057, pg 14)

= (50.7*0.0045)/(201*10³

= 11.34 N/mm² (Z_y= I_z/(y/2)

Warping Stress:

Warping Normal Stress: σ_wf=- E*W_ns*φ” ( CL 2.1.5 of P057, pg 10)

=-205000*10053.3*-7.2146*10^^-8 = 148.7 N/mm²

Warping Shear Stress: σ_tw= 0. (Warping Shear is not produced at end point of flange)

Pure Torsion Stress:

Shear Stress: = Gt Ф’ ( CL 2.1.4 of P057, pg 9)

= 78846.15 * 14.2 * 4.23*10^^-8 = 0.0475

Direct Shear Stress

Shear Stress due to Direct Shear V : τ_b = 0. (Shear Stress due to bending is not produced at the flange end)

The stress check will be performed using equation 6.1 of EN 1993-1-1:2005 as given below:

σ_xEd= Axial stress + (Bending Stress)_z +(Bending Stress)_y + Normal Stress due to warping.

τ_Ed =Shear stress due to Direct Shear + Shear stress due to pure torsion + Shear stress due to warping torsion.

σ_xEd = 86.82 + 11.34 + 148.7 = 246.86 N/mm²

τ_Ed = 0.0475

Hence, for point 1, the ratio is calculated as (246.86/275)² + 3(0.0475/275)² = 0.805.

Point 2:

Bending Stress:

σ_bz= 86.82 N/mm²

σ_byt= 11.34 N/mm²

Warping Stress:

Warping Normal Stress: σ_wf= 0 (Normal warping Stress is 0 at mid of flange)

Warping Shear Stress: σ_tw= - (E*W_ns*φ’’)/t = (-205000*734.486.10⁴*-9.295*10 ¹¹ / 14.2 )

= 9.86 N/mm²

Pure Torsion Stress:

Shear Stress: = 0.0475 (Same as point 1)

Direct Shear Stress

Shear Stress due to Direct Shear V : τ_b =(V_y*Q_f) /(I_z*t)

Where V_y is the shear at the section and Q _f is the statical moment at the flange.

Q _f= (205.8*14.2*97.7)/2 = = 142757 mm³

Hence,

τ_b= (50.7*10³*142757) / (6120*10⁴*14.2) = 8.32 N/mm²

σ_xEd= 86.82 + 11.34 = 98.16 N/mm²

τ_Ed = 8.32+9.86+0.0475 = 18.23 N/mm²

Hence, for point 2, the ratio is calculated as (98.16/275)² + 3(18.23/275)² = 0.14.

Point 3:

Stress at point 3 will be same as in point 2.

Point 4:

Bending Stress:

σ_bz= 0 ( At mid web, bending stress is always 0)

σ_byt= 11.34 N/mm²

Warping Stress:

Warping Normal Stress: σ_wf= 0 (Normal warping Stress is 0 at web, warping is always at flange)

Warping Shear Stress: σ_tw= 0 (Warping Shear Stress is 0 at web, warping is always at flange)

Pure Torsion Stress:

Shear Stress: τ_tw= 0 ( Shear stress at mid web is 0 due to pure torsion)

Direct Shear Stress

Shear Stress due to Direct Shear V: τ_b =(V_y*Q_w) /(I_z*t)

Where V_y is the shear at the section and Q _w is the statical moment at the web.

Q _w= (205.8*14.2*97.7) + (90.6*9.2*90.6)/2

= 323273 mm³

Hence, τ_{b =}(50.7*10³*323273)/(6120*10⁴*9.2) = 29.11 N/mm²

σ_xEd = 11.34 N/mm²

τ_Ed= 29.11 N/mm²

Hence, the ratio for point 4 is calculated as (11.34/275)² + 3(29.11/275)² = 0.035.

Hence critical ratio is 0.805 at location 1 of the cross section. The same ratio is reported by Staad. An extract from the STAAD Output File is as below:

D. Validation of the ratio calculated for ratio 6.2.7(9):

As per clause 6.2.7(9) of the code, for combined shear and Torsional moment the plastic shear resistance accounting for Torsional effects should be reduced from Vpl,Rd to V pl,T,Rd and the design shear force should be less than this reduced shear strength. The critical ratio for this clause is calculated at a distance of 1 metre from the start of the section.

Shear Area: A_v= A-2bt_f+ (t_w+2r) +t_f

= 7640-2*(205.8*14.2)+(9.2+2*10.2)*14.2

= 2215.6 mm²

Hence, V _pl,Rd= (2215.6*275)/√3

= 351.77 KN

And,

τ _t, Ed = Gt Ф’ = 78846.15*14.2*3.347*10^-5

= 37.48 Nmm (Critical Section is 1 m from start of member, hence Ф’ corresponding to z =1m is used)

Hence, V _{pl, T,Rd}= √(1-37.48/(1.25 275/√3)) )*351.7 = 316.6 KN = 316.6 KN

V _Edat 1m is 51KN (from Staad)

So,

( V _Ed ) / (V _pl,T,Rd ) = 51/316.6 = 0.16 (against the value reported by Staad as 0.155)

An extract from the output file shows the ratio reported as below:

E. Torsion Check as per CL A.2 of EN 1993-6:2007

This clause is essentially for crane run-away beams and found in EN 1993-6:2007, Annex A. However, this is checked for I sections subjected to bending and torsion in order to assess the interaction between torsion and lateral torsional buckling.

Clause A.1 states that members that are subjected to combined bending and torsion should satisfy: